如何克隆一个类但不复制可变属性？

Question

我正在尝试编写能够跟踪创建的类实例并通过会话保存它的 python 代码。我试图通过在类减速中创建一个列表来跟踪实例。我的代码如下：

class test_object:
    _tracking = []

    def __init__(self, text):
        self.name = text
        test_object._tracking.insert(0, self)

with open("tst.pkl", mode="rb") as f:
    try:
        pickles = dill.load(f)
    except:
        pickles = test_object
        logger.warning("Found dill to be empty")
f.close()

当腌制数据为空时，我的问题是处理。在这种情况下，我想做的是简单地使用基类。我遇到的问题是test_object._tracking 最终等于pickles._tracking。有没有办法复制test_object，这样当test_object._tracking 获得更新时，pickles._tracking 保持不变？

Answer 1

您可以执行以下操作：

import dill 

class test_object:
    _tracking = []

    def __init__(self, text):
        self.name = text
        test_object._tracking.insert(0, self)

test_1 = test_object("abc")
print(test_object._tracking)
# result: [<__main__.test_object object at 0x11a8cda50>]

with open("./my_file.txt", mode="rb") as f:
    try:
        pickles = dill.load(f)
    except:
        pickles = type('test_object_copy', test_object.__bases__, dict(test_object.__dict__))
        pickles._tracking = []

        print("Found dill to be empty")
# The above results in "Found dill to be empty"

print(pickles._tracking)
# prints []

它会将泡菜设置为原始类的副本。然后它的跟踪属性将为空，并且与原始的“跟踪”不同。